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Kinetic pressure

by Kenny Lee
Tags: kinetic, pressure
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Kenny Lee
#1
Nov3-05, 04:38 AM
P: 76
Please visit this site for me. I'm having a little trouble with the explanation.
http://hyperphysics.phy-astr.gsu.edu...kinthe.html#c3
Note however that you can find a similar explanation in all textbooks, so I'm in no way questioning the validity of the information.

The average force that the wall exerts on the molecule is impulse/ time (N's second law). But the time they have made use of, is the time for a 'round trip'. Shouldn't it be the time taken for the collision?
Thans
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Nov3-05, 06:55 AM
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Quote Quote by Kenny Lee
Please visit this site for me. I'm having a little trouble with the explanation.
http://hyperphysics.phy-astr.gsu.edu...kinthe.html#c3
Note however that you can find a similar explanation in all textbooks, so I'm in no way questioning the validity of the information.
The average force that the wall exerts on the molecule is impulse/ time (N's second law). But the time they have made use of, is the time for a 'round trip'. Shouldn't it be the time taken for the collision?
Thans
This is an "average" calculation, assuming there are enough molecules that some are striking the wall at any given time. In terms of individual molecules, the force that one molecule contributes is, say, F during the collision, 0 while it is making the trip to the other wall and back. It's average contribution is the impulse divide by the time of the entire trip. You can, if you like, think of it as "total force"- the actual force times the duration of the collision plus 0 times the duration of the trip- divided by the total time.
Kenny Lee
#3
Nov3-05, 07:08 AM
P: 76
But couldn't we then make use of the 'average' over a longer period; lets say two 'round trips'. In which case, we would obtain a duration of 4d/v. And a different final result.
Or have I misunderstood?


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